Question

a). A prototype car starts from rest and accelerates at a constant rate to 30 m/s in 5.6 seconds. How much distance did the car travel during the 5.6-second accelerated motion?

b). A car was parked on the side of a hillside road at the top of a hill. Unbeknown to the owner, its parking brake failed and tremors from nearby construction set it in motion with negligible initial speed down the hill. Assume the acceleration was constant down the hill, the car was given a speed ticket by a traffic camera at the foot of the hill, which measured the speed of the car at 30 m/s. The video recording showed the car took 38 to slide down the hill. How long is the road on the side of the hill?

c). At time 3 s, the speedometer of your car reports 37 m/s. At a later time 10, your car reports 34 m/s. Assuming you are accelerating at a constant rate. Find your acceleration.

d). Starting from rest, it took you 11 seconds to accelerate your car to 32 m/s at a constant acceleration. Find your acceleration.

e). A speeding driver noticed a police car parked about 436 m in from of him on the side of the road and reduced his speed at a constant rate for a duration of 10.3 seconds as he drove past the police car, at the legal speed of 31 m/s. What was the car's acceleration (include a negative sign in your answer)? Hint: the initial position of the car can be set at the origin.

Answer 1

a) The car starts from rest. So the initial velocity u = 0. The final velocity of the car is v = 30m/s and the time taken for it is t = 5.6s. So the acceleration of the car is given by,

So the distance travelled by the car is,

So the car travelled 84.04m.

b) The car was at rest. So the initial velocity u = 0. Velocity of the car at the foot of the hill is v = 30m/s and the time taken for it is t = 38s. So the acceleration of the car is,

The length of the road is given by,

So the road on the side of the hill is 570.38m.

c) At t = 3s, the speed of the car was u = 37m/s. At t = 10s, the speed of the car was v = 34m/s. So the time taken for the car to change the speed from u to v is t = 7s. So the acceleration of the car was

So the car is decelerating at 0.43m/s2.

d) The car was at rest. So the initial velocity u = 0. The final velocity of the car is v = 32m/s and the time taken for it is t = 11s. So the acceleration of the car is,

The acceleration of the car is 2.91m/s2.

e) The distance travelled by the car is S = 436m. The time taken to reduce the speed is t = 10.3s. The final velocity of the car is v = 31m/s. Since it is a decelerated motion, we can write the equation of motion as

Now the distance covered is,

So the initial velocity of the car is 53.66m/s.

Now the acceleration of the car is,

Since the motion is decelerated the acceleration of the car is a = -2.2m/s2.